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1 e’2x
ln 8 e2x
= ++ dx
4 2 4
1
1 ln 8 x
e + e’x dx
=
21
1
= [ex ’ e’x ]ln 8
1
2
63 e 1
= ’+.
16 2 2e
You Try It: Set up, but do not evaluate, the integral for the arc length of the graph

of y = sin x on the interval π/4 ¤ x ¤ 3π/4.
Sometimes an arc length problem is more conveniently solved if we think of the
curve as being the graph of x = g(y). Here is an example.
243
CHAPTER 8 Applications of the Integral

EXAMPLE 8.22
Calculate the length of that portion of the graph of the curve 16x 2 = 9y 3
between the points (0, 0) and (6, 4).

SOLUTION
We express the curve as
3
x = y 3/2 , 0 ¤ y ¤ 4.
4
Then dx/dy = 9 y 1/2 . Now, reversing the roles of x and y in ( ), we ¬nd that
8
the requested length is
4 4
1 + [(9/8)y 1/2 ]2 dy = 1 + (81/64)y dy.
0 0

This integral is easily evaluated and we see that it has value [2 · (97)3/2 ’ 128]/
243.
Notice that the last example would have been considerably more dif¬cult (the
integral would have been harder to evaluate) had we expressed the curve in the
form y = f (x).

You Try It: Write the integral that represents the length of a semi-circle and
evaluate it.

8.5.2 SURFACE AREA
Let f (x) be a non-negative function on the interval [a, b]. Imagine rotating the
graph of f about the x-axis. This procedure will generate a surface of revolution,
as shown in Fig. 8.31. We will develop a procedure for determining the area of such
a surface.



y = f (x)




a b



Fig. 8.31
244 CHAPTER 8 Applications of the Integral

We partition the interval [a, b]:
a = x0 ¤ x1 ¤ x2 ¤ · · · ¤ xk’1 ¤ xk = b.
Corresponding to each pair of elements xj ’1 , xj in the partition is a portion of curve,


y = f (x)




a b




Fig. 8.32

as shown in Fig. 8.32. When that piece of curve is rotated about the x-axis, we obtain
a cylindrical surface. Now the area of a true right circular cylinder is 2π · r · h. We
do not have a true cylinder, so we proceed as follows. We may approximate the
radius by f (xj ). And the height of the cylinder can be approximated by the length
of the curve spanning the pair xj ’1 , xj . This length was determined above to be
about
1/2
1 + [f (xj )]2 xj .
Thus the area contribution of this cylindrical increment of our surface is about
1/2
2π · f (xj ) 1 + [f (xj )]2 xj .
See Fig. 8.33. If we sum up the area contribution from each subinterval of the
partition we obtain that the area of our surface of revolution is about
k
1/2
2π · f (xj ) 1 + [f (xj )]2 (—)
xj .
j =1

But this sum is also a Riemann sum for the integral
b
1/2
f (x) 1 + [f (x)]2
2π dx.
a

As the mesh of the partition gets ¬ner, the sum (—) more closely approximates
what we think of as the area of the surface, but it also converges to the integral.
245
CHAPTER 8 Applications of the Integral




,x
Fig. 8.33


We conclude that the integral
b
f (x)(1 + [f (x)]2 )1/2 dx

a
represents the area of the surface of revolution.
EXAMPLE 8.23
Let f (x) = 2x 3 . For 1 ¤ x ¤ 2 we rotate the graph of f about the x-axis.
Calculate the resulting surface area.

SOLUTION
According to our de¬nition, the area is
2
f (x)(1 + [f (x)]2 )1/2 dx

1
2
= 2π 2x 3 (1 + [6x 2 ]2 )1/2 dx
1
2 3
π
= (1 + 36x 4 )1/2 (144x 3 ) dx.
54 2
1

This integral is easily calculated using the u-substitution u = 36x 4 , du =
144x 3 dx. With this substitution the limits of integration become 36 and 576;
the area is thus equal to
576 3 576
π π
(1 + u)1/2 du = (1 + u)3/2
54 2 54 36
36
π
= [(577)3/2 ’ (37)3/2 ]
54
≈ 793.24866.
246 CHAPTER 8 Applications of the Integral

EXAMPLE 8.24
Find the surface area of a right circular cone with base of radius 4 and
height 8.

SOLUTION
It is convenient to think of such a cone as the surface obtained by rotating
the graph of f (x) = x/2, 0 ¤ x ¤ 8, about the x-axis (Fig. 8.34). According
to our de¬nition, the surface area of the cone is

8x 58
[1 + (1/2) ] dx = 2π
2 1/2
2π x dx
02 40

= 16 5π.

y




x




Fig. 8.34


You Try It: The standard formula for the surface area of a cone is

S = π r h2 + r 2 .
Derive this formula by the method of Example 8.24.
We may also consider the area of a surface obtained by rotating the graph of a
function about the y-axis. We do so by using y as the independent variable. Here
is an example:
EXAMPLE 8.25
Set up, but do not evaluate, the integral for ¬nding the area of the surface
obtained when the graph of f (x) = x 6 , 1 ¤ x ¤ 4, is rotated about the
y-axis.
247
CHAPTER 8 Applications of the Integral

SOLUTION
We think of the curve as the graph of φ(y) = y 1/6 , 1 ¤ y ¤ 4096. Then the
formula for surface area is
4096
1/2
φ(y) 1 + [φ (y)]2
2π dy.
1
Calculating φ (y) and substituting, we ¬nd that the desired surface area is the
value of the integral
4096
y 1/6 1 + (1/6)y ’5/6
2 1/2
2π dy.
1

You Try It: Write the integral that represents the surface area of a hemisphere of
radius one and evaluate it.



8.6 Hydrostatic Pressure
If a liquid sits in a tank, then it exerts force on the side of the tank. This force is
caused by gravity, and the greater the depth of the liquid then the greater the force.
Pascal™s principle asserts that the force exerted by a body of water depends on
depth alone, and is the same in all directions. Thus the force on a point in the side
of the tank is de¬ned to be the depth of the liquid at that point times the density of
the liquid. Naturally, if we want to design tanks which will not burst their seams, it
is important to be able to calculate this force precisely.




Fig. 8.35

Imagine a tank of liquid having density ρ pounds per cubic foot as shown in
Fig. 8.35. We want to calculate the force on one ¬‚at side wall of the tank. Thus
we will use the independent variable h to denote depth, measured down from the
surface of the water, and calculate the force on the wall of the tank between depths
h = a and h = b (Fig. 8.36). We partition the interval [a, b]:
a = h0 ¤ h1 ¤ h2 ¤ · · · ¤ hk’1 ¤ hk = b.
248 CHAPTER 8 Applications of the Integral

Assume that the width of the tank at depth h is w(h). The portion of the wall between
h = hj ’1 and h = hj is then approximated by a rectangle Rj of length w(hj ) and
width h = hj ’ hj ’1 (Fig. 8.37).




h=a
h=b




Fig. 8.36


w(hj)




Y ,h
FL
Fig. 8.37
AM

Now we have the following data:
Area of Rectangle = w(hj ) · h square feet
TE



Depth of Water ≈ hj feet
Density of Liquid = ρ pounds per cubic foot.
It follows that the force exerted on this thin portion of the wall is about
Pj = hj · ρ · w(hj ) · h.
Adding up the force on each Rj gives a total force of
k k
Pj = hj ρ w(hj ) h.
j =1 j =1

But this last expression is a Riemann sum for the integral
b
(—)
hρw(h) dh.
a

EXAMPLE 8.26
A swimming pool is rectangular in shape, with vertical sides. The bottom
of the pool has dimensions 10 feet by 20 feet and the depth of the water
249
CHAPTER 8 Applications of the Integral

is 8 feet. Refer to Fig. 8.38. The pool is full. Calculate the total force on one
of the long sides of the pool.


20
10
8




Fig. 8.38


SOLUTION
We let the independent variable h denote depth, measured vertically down
from the surface of the water. Since the pool is rectangular with vertical sides,
w(h) is constantly equal to 20 (because we are interested in the long side). We
use 62.4 pounds per cubic foot for the density of water. According to (—), the
total force on the long side is
8 8
h · 62.4 · w(h) dh = h · 62.4 · 20 dh = 39936 lbs.
0 0

You Try It: A tank full of water is in the shape of a cube of side 10 feet. How
much force is exerted against one wall of the tank between the depths of 3 feet and
6 feet?
EXAMPLE 8.27
A tank has vertical cross section in the shape of an inverted isosceles tri-
angle with horizontal base, as shown in Fig. 8.39. Notice that the base of the
tank has length 4 feet and the height is 9 feet. The tank is ¬lled with water to
a depth of 5 feet. Water has density 62.4 pounds per cubic foot. Calculate
the total force on one end of the tank.


4 ft


9 ft

5 ft


Fig. 8.39
250 CHAPTER 8 Applications of the Integral

SOLUTION
As shown in Fig. 8.40, at depth h (measured down from the surface of the
water), the tank has width corresponding to the base of an isosceles triangle
similar to the triangle describing the end of the tank. The height of this triangle
is 5 ’ h. Thus we can solve
4
w(h)
=.
5’h 9
We ¬nd that
4
w(h) = (5 ’ h).
9
According to (—), the total force on the side is then
5 4
h · 62.4 · (5 ’ h) dh ≈ 577.778 lbs.
9
0


4


9

5
5_h


Fig. 8.40


EXAMPLE 8.28
An aquarium tank is ¬lled with a mixture of water and algicide to keep the
liquid clear for viewing. The liquid has a density of 50 pounds per cubic foot.
For viewing purposes, a window is located in the side of the tank, with center
20 feet below the surface. The window is in the shape of a square of side

4 2 feet with vertical and horizontal diagonals (see Fig. 8.41). What is the
total force on this window?

SOLUTION
As usual, we measure depth downward from the surface with independent
variable h. Of course the square window has diagonal 4 feet. Then the range of
integration will be h = 20’4 = 16 to h = 20+4 = 24. Refer to Fig. 8.42. For
h between 16 and 20, we notice that the right triangle in Fig. 8.42 is isosceles
and hence has base of length h ’ 16. Therefore
w(h) = 2(h ’ 16) = 2h ’ 32.
251
CHAPTER 8 Applications of the Integral


4√2




Fig. 8.41

h = 16

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